# Significant Figure Help

1. Feb 10, 2005

### tony873004

My teacher takes off points for significant figure mistakes, but advises us to keep a few extra significant figures until the end. How would I do this problem:

Compute the velocity of the Sun around the Milky Way. Assume a circular orbit. Use the values given by Croswell for distance from galactic center and orbital period. Use unit conversions in Appendix A in Hester.

Distance of Sun to Center of Galaxy as given by Croswell = 27,000 ly
Period of the Sun’s orbit around the Galaxy as given by Croswell = 230,000,000 years
1 ly = 9.4605 * 10^15 m as given by Hester
1 yr = 3.15581 * 10^7s as given by Hester

Distance of Sun to Center of Galaxy = 9.4605 * 10^15 * 27,000 = 2.554335 * 10^20 m
Circumference of Sun’s orbit around the galaxy = 2 π r
Circumference of Sun’s orbit around the galaxy = 2 π 2. 554335 * 10^20 m
Circumference of Sun’s orbit around the galaxy = 1.60493601416146 * 10^21 m

Period of Sun’s orbit in seconds = s/yr * Pyears
Period of Sun’s orbit in seconds = 3.15581 * 10^7s/yr * 2.3 * 10^8 yr
Period of Sun’s orbit in seconds = 7.258363 * 10^15 s

V = d/t
V = 1.60493601416146 * 10^21 m / 7.258363 * 10^15 s
V =221115.424257709 m / s
V = 221.115424257709 km / s

I haven't dumped any figures yet, simply pasting the results out of my calculator. But the 27000 light years confuses me. Is this only 2 significant figures, since I imagine the center of the galaxy isn't exactly 27000 ly, but probably somewhere inbetwen 26500 and 27500. And what if it were known that the center of the galaxy was exactly 27000 light years away. How would that be indicated? 27000.0?

And should I be dumping insignificant figures in the intermediate steps? Or only for the final answer?
If you were a teacher that marked off for significant figures, how would you want me to do the above problem?

2. Feb 10, 2005

### dextercioby

It woul be fair if you followed the same line from beginning to the end.So if you had 2 significant digits to begin with,i see no reason to end up with 15 or so.
I guess it would all come down to approx.220Km/s

Daniel.

3. Feb 10, 2005

### Staff: Mentor

27000 has only two significant figures; if it had 5 sig figs, it would be written like this: 27.000 x 10^3. (Or 2700.0 x 10^1, etc.) 27000.0 has 6 sig figs.

If you need to show your intermediate results, you should ask your prof what he wants to see. Otherwise, let the calculator keep all the digits it wants and round off the final answer to the proper number of sig figs.

4. Feb 10, 2005

### tony873004

Thanks Dex, and Doc. Another question:

Compute the perihelion and aphelion distances for each of the planets in AU.

Periapsis = A (1 - e)
Apoapsis = A (1 + e)

Planet b (A = 0.115 AU, e = 0.02)
Periapsis = 0.113 AU
Apoapsis = 0.117 AU

The e=0.02 number, is that only 1 significant figure? Or does the 0 after the decimal point count? I'm guessing that its 2 significant figures. But how about if it were written as 2 * 10^-2. Would that be only 1 significant figure.

So if my guess is right, then my answers should be:
Periapsis = 0.11 AU
Apoapsis = 0.12 AU

also, how about a number like 0.24. Is that only 2 significant digits, or does the 0 left of the decimal count and make it 3 significant digits?
Thanks...

Last edited: Feb 10, 2005
5. Feb 10, 2005

### dextercioby

In any form,the no.0.02 has one sign.figure,namely 2.As Doc said,maybe it's better if you keep all intermediary figures and round only at the end.In that case,it would be no problem with 0.24.Depending on the no.of significant digits,it would be either 0.24 (2 digits) or 0.2 (one digit).

Daniel.

6. Feb 10, 2005

### Staff: Mentor

Careful. Taken alone 0.02 has 1 sig fig. (Zeroes to the left don't count.) But when adding or subtracting, it's the number of decimal places that determine the digits in the answer. So 3 + 0.02 = 3 (1 sig figs), but 3.03 + 0.02 = 3.05 (3 sig figs!). But the 1 in 1 + e is a pure number = 1.000000... (as many digits as you like; it's exactly 1), so 1.00 + 0.02 = 1.02 (3 sig figs).
0.24 has 2 sig figs, not 3.

7. Feb 10, 2005

### tony873004

Here's what doesn't make sense to me. Take the example

Periapsis = A (1 - e)
Planet b (A = 0.115 AU, e = 0.02)

Periapsis = 0.1127 (calculator number)

0.02 only has one significant digit, so I'd have to write my answer as

Periapsis = 0.1

But the whole purpose of significant digits is because 0.02 could represent any number from 0.015 to 0.025, and 0.115 can represent any number from 0.1145 to 0.1155. But even if we use these extremes:

Periapsis = 0.113275 (calculator number using e = 0.015, A = 0.115)
Periapsis = 0.112125 (calculator number using e = 0.025, A = 0.115)

Periapsis = 0.1127825 (calculator number using e = 0.015 A = 0.1145)
Perapsis = 0.1137675 (calculator number using = 0.015, A = 0.1155)

Perapsis = 0.1116375 (calculator number using e=0.025, A = 0.1145)
Perapsis = 0.1126125 (calculator number using e=0.025, A = 0.1155)

The final answer doesn't start to fall apart until the 3rd digit to the right of the decimal.
So it seems to me like I'm throwing away a perfectly good 1 by writing 0.1 instead of 0.11.

8. Feb 10, 2005

### dextercioby

I'll have to repeat myself,but it all depends on the requirements of the problem.If the problem asks for one signif.figure only,then u must provide 0.1,even if u know that 0.11 is an answer closer to the real value.

Daniel.

9. Feb 10, 2005

### Staff: Mentor

This is incorrect. (Did you read my last post?) A has 3 sig figs; (1 - e) = 0.98 has 2 sig figs. Thus A (1 - e) will have 2 sig figs.

This may help you understand the rules for arithmetic with sig figs: http://en.wikipedia.org/wiki/Significance_arithmetic

Last edited: Feb 10, 2005
10. Feb 10, 2005

### tony873004

Now I get it. Thanks, Doc! I did read your previous post. It just didn't sink in until now.